# stacks/stacks-project

Confusing proof made less confusing

Hopefully

Thanks to Noah Olander
https://stacks.math.columbia.edu/tag/00Q9#comment-4514
https://stacks.math.columbia.edu/tag/00Q9#comment-4516
 The case $n = 1$. We may replace $R$ by its integral closure in $S$ (Lemma \ref{lemma-quasi-finite-permanence} guarantees that $R \to S$ is still quasi-finite at $\mathfrak q$). Thus we may assume $R \subset S$ is integrally closed in $S$. $R \subset S$ is integrally closed in $S$, in other words $R = S'$. Consider the map $\varphi : R[x] \to S$, $x \mapsto x_1$. (We will see that $\varphi$ is not injective below.) By assumption $\varphi$ is finite. Hence we are in Situation Thus there exists an element $s \in J$, $s \not\in \mathfrak q$. By definition of $J$ we may write $s = \varphi(f)$ for some polynomial $f \in R[x]$. Now let $I = \Ker(R[x] \to S)$. Since $\varphi(f) \in J$ Let $I = \Ker(\varphi : R[x] \to S)$. Since $\varphi(f) \in J$ we get $(R[x]/I)_f \cong S_{\varphi(f)}$. Also $s \not \in \mathfrak q$ means that $f \not \in \varphi^{-1}(\mathfrak q)$. Thus $\varphi^{-1}(\mathfrak q)$ is a prime of $R[x]/I$ at which $R \to R[x]/I$ is quasi-finite, see Lemma \ref{lemma-quasi-finite-local}. Let $C \subset R[x]/I$ be the integral closure of $R$. By Lemma \ref{lemma-quasi-finite-local}. Note that $R$ is integrally closed in $R[x]/I$ since $R$ is integrally closed in $S$. By Lemma \ref{lemma-quasi-finite-monogenic} there exists an element $h \in C$, $h \not \in \varphi^{-1}(\mathfrak q)$ such that $C_h \cong (R[x]/I)_h$. We conclude that there exists an element $h \in R$, $h \not \in R \cap \mathfrak q$ such that $R_h \cong (R[x]/I)_h$. Thus $(R[x]/I)_{fh} = S_{\varphi(fh)}$ is isomorphic to a principal localization $C_{h'}$ of $C$ for some $h' \in C$, $h' \not \in \varphi^{-1}(\mathfrak q)$. Since $\varphi(C) \subset S'$ we get $g = \varphi(h') \in S'$, $g \not \in \mathfrak q$ and moreover the injective map $S'_g \to S_g$ is also surjective because by our choice of $h'$ the map $C_{h'} \to S_g$ is surjective. localization $R_{h'}$ of $R$ for some $h' \in R$, $h' \not \in \mathfrak q$. \medskip\noindent The case $n > 1$. Consider the subring $R' \subset S$ which is the integral closure of $R[x_1, \ldots, x_{n-1}]$ in $S$. By Lemma \ref{lemma-four-rings} the extension in $S$. By Lemma \ref{lemma-quasi-finite-permanence} the extension $S/R'$ is quasi-finite at $\mathfrak q$. Also, note that $S$ is finite over $R'[x_n]$. By the case $n = 1$ above, there exists a $g' \in R'$,